$\overline{AB}$ = $5$ $\overline{AC} = {?}$ $A$ $C$ $B$ $5$ $?$ $ \sin( \angle ABC ) = \frac{4}{5}, \cos( \angle ABC ) = \frac{3}{5}, \tan( \angle ABC ) = \dfrac{4}{3}$
$\overline{AB}$ is the hypotenuse $\overline{AC}$ is opposite to $\angle ABC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle ABC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{5} $ $ \overline{AC}=5 \cdot \sin( \angle ABC ) = 5 \cdot \frac{4}{5} = 4$